410 lines
20 KiB
TeX
410 lines
20 KiB
TeX
\section{The Stabilizer Formalism and VOP-Free Graph States}
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\subsection{Stabilizers and Stabilizer States}
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This chapter discusses the stabilizer formalism that was introduced by Gottesman\cite{gottesman1997}
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for quantum error correction but soon proved to be a useful tool to describe a subset of states:
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the stabilizer states which can be simulated in polynomial time \cite{gottesman2008}.
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\begin{definition}
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\begin{equation}
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p \in P_n \Rightarrow p = \bigotimes\limits_{i=0}^n p_i \\
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\forall i: p_i \in P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\}
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\end{equation}
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Where $X = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right)$,
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$Y = \left(\begin{array}{cc} 0 & i \\ -i & 0\end{array}\right)$ and
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$Z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right)$ are the Pauli matrices and
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$I$ is the identity.
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$p \in P_n$ is called a multi-local Pauli operator.
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\end{definition}
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\begin{definition}
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For a group $G$, $g_1, ..., g_n$ are called the generators of $G$ iff $\forall g \in G: g = \prod\limits_{i \in J} g_i$ for a
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subsed $J$ of $I = \{1, ..., n\}$. We write $G = \langle g_i \rangle_i$ if G is generated by the $g_i$. The generators $g_i$ are chosen
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to be the smallest set of generators of $G$.
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\end{definition}
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The notation $\langle g_i \rangle_i \equiv \langle g_i \rangle_{i \in I}$ is used used as a shorthand
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notation for $\langle \{g_i\}_{i \in I} \rangle$.
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\begin{definition}
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\label{def:stabilizer}
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For a $n$ qbit state $\ket{\psi}$ $\langle S^{(i)} \rangle_{i}$ is called the stabilizer of $\ket{\psi}$ if
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\begin{enumerate}
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\item{$\forall i = 1, ..., n$ $S^{(i)} \in P_n$}
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\item{$\forall i,j = 1, ..., n$ $[S^{(i)}, S^{(j)}] = 0$ $S^{(i)}$ and $S^{(j)}$ commute}
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\item{$\forall i = 1, ..., n$ $S^{(i)}\ket{\psi} = +1 \ket{\psi}$}
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\end{enumerate}
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\end{definition}
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\begin{lemma}
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For every $\langle S^{(i)} \rangle_i$ fulfilling the first two conditions in definition \ref{def:stabilizer} there exists
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a (up to a global phase) unique state $\ket{\psi}$ fulfilling the third condition. This state is called stabilizer state.
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\end{lemma}
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\begin{proof}
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All multi-local Pauli operators are hermitian (observables in terms of quantum mechanics), as the $S^{(i)}$
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commute they have a common set of eigenstates. Because each $S^{(i)}$ has eigenvalues $+1, -1$, there
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exist $2^n$ eigenstates, one state $\ket{\psi}$ with eigenvalue $+1$ for all $S^{(i)}$. As the dimension of $n$ qbits is $2^n$
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the state $\ket{psi}$ is unique up to a global phase.
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\end{proof}
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One can study the dynamics of stabilizer states using only the stabilizers\cite{nielsen_chuang_2010}. Two important
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cases are the unitary transformation of a state and the measurement of a qbit. When applying a unitary gate to a stabilizer
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state $\ket{\psi}$ the resulting state will in general be no stabilizer state anymore, however there exists a group of
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transformations that map stabilizers to other stabilizers: the Clifford group.
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\begin{definition}
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\begin{equation}
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C_n := \{U \in SU(2) | UpU^\dagger \in P_n \forall p \in P_n\}
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\end{equation}
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is called the Clifford group on $n$ qbits.
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$C_1$ is called the local Clifford group.
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\end{definition}
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The properties of this group will be discussed later, for the time being is existence is enough.
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\begin{lemma}
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Let $\ket{\psi}$ be stabilized by $\langle S^{(i)} \rangle_i$, then $U\ket{\psi}$ is stabilized
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by $\langle US^{(i)}U^\dagger \rangle_i$.
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\end{lemma}
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\begin{proof}
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$$ U\ket{\psi} = US^{(i)}\ket{\psi} = US^{(i)}U^\dagger U\ket{\psi}$$
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So $U\ket{\psi}$ is a $+1$ eigenstate of $US^{(i)}U^\dagger$.
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\end{proof}
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This is an important insight that is used for simulations\cite{gottesman_aaronson2008}, as
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updating the $n$ stabilizers that are a tensor product of $n$ Pauli matrices scales with roughly
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$\mathcal{O}(n^2)$ instead of $\mathcal{O}(2^n)$ for the state vector approach.
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Every qbit can be measured in the $X, Y$ or $Z$ basis which is a projection using
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\begin{equation}
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\frac{I + (-1)^s g_a}{2}
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\end{equation}
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Where $g_a \in \{Z_a, Y_a, X_a\}$ and $s \in \{0, 1\}$. How the stabilizers change when
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measuring a qbit is given by the following lemma:
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\begin{lemma}
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\label{lemma:stab_measurement}
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Let $J := \{ S^{(i)} | [g_a, S^{(i)}] \neq 0\}$. Then
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\begin{enumerate}
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\item{If $J = \{\}$, one value is measured with probability $1$ and the stabilizers are unchanged.}
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\item{If $J \neq \{\}$, $1$ and $0$ are measured with probability $\frac{1}{2}$ and after choosing
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a $j \in J$ the new state $\ket{\psi'}$ is stabilized by
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\begin{equation}
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\langle \{(-1)^s g_a\} \cup \{S^{(i)} S^{(j)} | S^{(i)} \in J \setminus \{S^{(j)}\} \} \cup J^c\rangle
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\end{equation}}
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\begin{enumerate}
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\item{As $g_a$ commutes with all stabilizers, $\ket{\psi}$ is an eigenstate of $g_a$,
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so the result of the measurement is deterministic and $\ket{\psi}$ is left unchanged.}
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\item{As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators,
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$S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
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\begin{equation}
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\begin{aligned}
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P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\
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&= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\
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&= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
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&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\
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&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
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&= P(s=-1)
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\end{aligned}
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\notag
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\end{equation}
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With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$.
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Further for $S^{(i)},S^{(j)} \in J$
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\begin{equation}
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\begin{aligned}
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\frac{I + (-1)^sg_a}{2}\ket{\psi} &= \frac{I + (-1)^sg_a}{2}S^{(j)}S^{(i)} \ket{\psi} \\
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&= S^{(j)}\frac{I + (-1)^{s + 1}g_a}{2}S^{(i)} \ket{\psi} \\
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&= S^{(j)}S^{(i)}\frac{I + (-1)^{s + 2}g_a}{2}\ket{\psi} \\
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&= S^{(j)}S^{(i)}\frac{I + (-1)^sg_a}{2}\ket{\psi}
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\end{aligned}
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\notag
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\end{equation}
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the state after measurement is stabilized by $S^{(j)}S^{(i)} i,j \in J$, and by
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$S^{(i)} \in J^c\setminus\{a\}$. $g_a$ trivially stabilizes $\ket{\psi'}$.
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}
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\end{enumerate}
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\end{proof}
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\subsection{The Vertex Operator-Free Graph States}
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In order to understand some essential transformations of graph states it is necessary
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to study the vertex operator-free graph states first, partially because the graph states as used in this paper
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were derived from the vertex operator-free graph states.
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\begin{definition}
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\label{def:vop_free_g_state}
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A $n$ qbit vertex operator-free graph state $\ket{\overline{G}}$ is associated with a graph $\bar{G} = (V, E)$
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by the $n$ operators
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\begin{equation}
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K^{(i)}_G := X_i \left(\prod\limits_{\{i, j\} \in E} Z_j\right)
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\end{equation}
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for all $i \in V$ where for some operator $O$ $O_i$ indicates that it acts on the $i$-th qbit.
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$\ket{\overline{G}}$ is the $+1$ eigenstate of all $n$ $K^{(i)}_G$.
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\end{definition}
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\begin{corrolary}
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All $K^{(i)}_G$ commute and are hermitian. Therefore they have a common set of eigenstates
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(in particular definition \ref{def:vop_free_g_state} is well defined).
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In terms of quantum mechanics $K^{(i)}_G$ are observables.
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Further as $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$ which are
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multi-local Pauli operators, $\{K^{(i)}_G | i \in \{0, ..., n-1\}\}$ is the stabilizer
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of $\ket{\overline{G}}$ and $\ket{\overline{G}}$ is a stabilizer state.
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\end{corrolary}
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\begin{proof}
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As $X_i$ and $Z_i$ are hermitian their product is hermitian.
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Consider the case $\{i,j\} \notin E$ first:
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\begin{equation}
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\begin{aligned}
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\left[K^{(i)}_G, K^{(j)}_G\right] = \left[X_i \prod\limits_{\{i, n\} \in E} Z_n, X_j \prod\limits_{\{j, m\} \in E} Z_m\right] = 0
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\end{aligned}
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\end{equation}
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As operators acting on different qbits commute. The case $\{i,j\} \in E$ is slightly less trivial:
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\begin{equation}
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\begin{aligned}
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\left[K^{(i)}_G, K^{(j)}_G\right] &= \left[X_i \left(\prod\limits_{\{i, n\} \in E, n \neq j} Z_n\right) Z_j, X_j \left(\prod\limits_{\{j, m\} \in E, m \neq i} Z_m\right) Z_i\right] \\
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&= \left[X_i Z_j \prod\limits_n Z_n, X_j Z_i \prod\limits_m Z_m\right]\\
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&= \left(X_i Z_j X_j Z_i - X_j Z_i X_i Z_j\right) \prod\limits_n Z_n \prod\limits_m Z_m \\
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&= \left(Z_j X_j X_i Z_i - X_j Z_j Z_i X_i\right) \prod\limits_n Z_n \prod\limits_m Z_m \\
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&= \left((-1)^2X_j Z_j Z_i X_i - X_j Z_j Z_i X_i\right) \prod\limits_n Z_n \prod\limits_m Z_m \\
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&= 0
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\end{aligned}
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\end{equation}
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as $X$, $Z$ anticommute.
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\end{proof}
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\begin{lemma}
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\begin{equation}
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\ket{\overline{G}} = \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right) \left(\prod\limits_{l \in V} H_l\right) \ket{0}
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\end{equation}
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In particular definitions \ref{def:vop_free_g_state} and \ref{def:graph_state} are consistent, when there are no
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vertex operators on the graph state $\ket{G}$.
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\end{lemma}
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\begin{proof}
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Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$ $X_i \ket{+} = +1 \ket{+}$.
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Set $\ket{\tilde{G}} := \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right)\ket{+}$.
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\begin{equation}
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\begin{aligned}
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K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,j\} \in E} CZ_{l,j} \right) \ket{+} \\
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& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\
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& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) X_i \ket{+} \\
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& = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{2\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\
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& = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\
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& = +1 \ket{\tilde{G}}
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\end{aligned}
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\end{equation}
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as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$.
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\end{proof}
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\begin{definition}
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Let $\bar{G} = (V, E)$ be a graph as in definition \ref{def:vop_free_g_state}.
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For a vertex $i \in V$ $n_i := \{j \in V | \{i,j\} \in E\}$ is called the neighbourhood
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of $i$.
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\end{definition}
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\begin{lemma}
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Let $\bar{G}$, $\ket{\bar{G}}$, $K_G^{(i)}$ be as in definition \ref{def:vop_free_g_state} and for
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a vertex $a \in V$ set
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\begin{equation}
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M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}
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\end{equation}
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Then the graph $\bar{G}'$ associated with $\ket{\bar{G}'} = M_a\ket{\bar{G}}$ is changed according to
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the following equations:
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\begin{equation}
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\begin{aligned}
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n_a' &= n_a \\
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n_j' &= n_j, \hbox{ if } j \notin n_a\\
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n_j' &= (n_j \cup n_a) \setminus (n_j \cap n_a), \hbox{ if } j \in n_a
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\end{aligned}
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\end{equation}
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I.e. the neighbourhood of $a$ is toggled.
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\end{lemma}
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\begin{proof}
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$\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger \rangle_i$, so it is sufficient
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to study how the $ K_G^{(i)}$ change under $M_a$.
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At first note that $[K_G^{(a)}, M_a] = 0$ and $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$,
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so the first two equations follow trivially. For $j \in n_a$ set
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\begin{equation}
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\begin{aligned}
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S^{(j)} &= M_a K_G^{(j)} M_a^\dagger \\
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&= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right)
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\sqrt{iZ_j} K_G^{(j)} \sqrt{iZ_j}^\dagger
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\left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right)
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\sqrt{-iX_a}^\dagger \\
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&= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right)\sqrt{iZ_j}
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X_j \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) Z_a
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\sqrt{iZ_j}^\dagger
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\left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right)
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\sqrt{-iX_a}^\dagger \\
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&= \sqrt{iZ_j} X_j\sqrt{iZ_j}^\dagger\left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)
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\sqrt{-iX_a} Z_a \sqrt{-iX_a}^\dagger \\
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&= (-1)^2 Y_j Y_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)\\
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&= (-1)i^2 Z_j X_a X_j Z_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)
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\end{aligned}
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\end{equation}
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One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenvalue
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of the new $K_{G'}^{(i)}$. It is clear that $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$.
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To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a = \{j\} \cup I$ and $n_j = \{a\} \cup J$.
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Then
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\begin{equation}
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\begin{aligned}
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S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in J} Z_l\\
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&= Z_j X_a X_j Z_a \left(\prod\limits_{l \in J} Z_l\right)
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\left(\prod\limits_{l \in I}Z_l\right)
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\left(\prod\limits_{l \in I}Z_l\right) \\
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&= Z_j X_a X_j Z_a \left(\prod\limits_{l \in ((I\cup J) \setminus (I\cap J))} Z_L\right)
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\left(\prod\limits_{l \in I}Z_l\right) \\
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&= K_{G'}^{(a)} K_{G'}^{(j)} \\
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&= K_{G}^{(a)} K_{G'}^{(j)}
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\end{aligned}
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\end{equation}
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Using this ine can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$:
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\begin{equation}
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\ket{\bar{G}'} = S^{(j)}\ket{\bar{G}'} = K_{G}^{(a)} K_{G'}^{(j)}\ket{\bar{G}'}
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= K_{G'}^{(j)}K_{G}^{(a)}\ket{\bar{G}'} = K_{G'}^{(j)}\ket{\bar{G}'}
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\end{equation}
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Because $\{K_G^{(i)} | i \notin n_a\} \cup \{S^{(i)} | i\in n_a\}$ and
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$\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}$ are both $n$ commuting
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multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$
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and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$
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$\langle\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}\rangle$
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are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given
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in the third equation.
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\end{proof}
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These insights can be used to understand how measurement works on the vop-free graph state \cite{nielsen_chuang_2010}:
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Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a Pauli operator $g \in \{Z_a, Y_a, X_a\}$
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that is to be measured.
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Recalling lemma \ref{lemma:stab_measurement} the following relations follow immideately:
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\begin{enumerate}
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\item{For $g = X_a$ if $a$ is an isolated qbit $+1$ is measured with probability $1$
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and the state $\ket{\bar{G}}$ is unchanged.}
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\item{For $g = X_a$ and $a$ is non-isolated, choose $b \in n_a$ and the new stabilizers are
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\begin{equation}
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\langle \{(-1)^sX_a\}
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\cup \{K_G^{(b)}K_G^{(i)} | i \in n_a \setminus \{b\}\}
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\cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
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\end{equation}}
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\item{For $g = Z_a$ the new stabilizers are
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\begin{equation}
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\langle \{(-1)^sZ_a\}
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\cup \{K_G^{(i)} | i \in V \setminus n_a\} \rangle
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\end{equation}}
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\item{For $g = Y_a$ the new stabilizers are
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\begin{equation}
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\langle \{(-1)^sY_a\}
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\cup \{K_G^{(a)}K_G^{(i)} | i \in n_a\}
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\cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
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\end{equation}}
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\end{enumerate}
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The states after the measurement are in general no vop-free graph states anymore,
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the following discussion will allow to construct new vop-free graph states and
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Clifford transformations from the vop-free graph state to the resulting state.
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In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged.
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\begin{lemma}
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\label{lemma:Z_measurement}
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\begin{enumerate}
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\item{For a result $+Z_a$ the new state is
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$\ket{+_Z}_a \otimes \ket{\bar{G}'}$
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with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for
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$i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$.
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}
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\item{For a result $-Z_a$ the new state is $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$
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with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$
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and $U = \prod\limits_{i \in n_a}Z_i$.}
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\begin{enumerate}
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\item{
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It is trivial that $Z_a$ and $K_G^{(i)} i \neq a$ stabilize $\ket{+_Z}_a \otimes \ket{\bar{G}'}$.
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$Z_aK_G^{(i)}$ for $i \in n_a$ do not act on $a$, so $\ket{\bar{G}'}$ is well-defined.
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}
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\item{
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The state $\ket{-_Z}_a \otimes \ket{\bar{G}'}$ is the $-1$ eigenstate of $K_G^{(i)}$ for all
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$i \in n_a$. This can be corrected by transforming $K_G^{(i)}$ to $-K_G^{(i)} = Z_i K_G^{(i)} Z_i^\dagger$
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which stabilizes the state $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$.
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}
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\end{enumerate}
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\end{proof}
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\begin{lemma}
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When $Y_a$ is measured with a result $s \in \{0, 1\}$ the state after the measurement is
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\begin{equation}
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\ket{(-1)^s_Y}_a \otimes U_{Y,s} \ket{\bar{G}'}
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\end{equation}
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\begin{equation}
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U_{Y,s} = \prod\limits_{l \in n_a} \sqrt{(-1)^s iZ_l}
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\end{equation}
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Where $\ket{\bar{G}'}$ is associated with the graph $G' = (V \setminus \{a\}, E')$
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with $E'$ being changed according to $\forall i \in n_a$ $n_i' = (n_i \cup n_a) \setminus (n_i \cap n_a) \setminus \{a\}$.
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\end{lemma}
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\begin{proof}
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It is known that $(-1)^s Y_a$ has to stabilize the state after measurement,
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further $\forall j \in n_a$ $S^{(j)} = K_G^{(a)} K_G^{(j)}$, are the stabilizers of the new
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state around the measured qbit and the stabilizers $K_G^{(j)}, j \notin n_a$ are unchanged.
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Let $j \in n_a$ and $n_a =: \{j\} \cup I$, $n_j =: \{a\} \cup J$.
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\begin{equation}
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\begin{aligned}
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S^{(i)} &= K_G^{(a)} K_G^{(j)} \\
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&= X_a \left(\prod\limits_{l \in n_a} Z_l\right)X_j\left(\prod\limits_{l\in n_j} Z_l\right) \\
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&= X_a Z_a Z_j X_j \left(\prod\limits_{l \in I} Z_l\right)\left(\prod\limits_{l \in J} Z_l\right)\\
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&= X_a Z_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)} Z_l\right) \\
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&= iY_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J) \setminus \{a\}} Z_l\right) \\
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&= (-1)^si S^{(a)} Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)\setminus\{a\}} Z_l\right) \\
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&= S^{(a)} \sqrt{(-1)^siZ_j} K_{G'}^{(j)} \sqrt{(-1)^siZ_j}^\dagger
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\end{aligned}
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\end{equation}
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with $G'$ as above, the rest of the argument is analogous to lemma \ref{lemma:Z_measurement}.
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\end{proof}
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